Question

One gram of a metal carbonate is neutralised completely by $200$ ml of $0.1N$ $HCl$. The equivalent weight of metal carbonate is:

• A. $25$
• B. $50$
• C. $100$
• D. $75$

Answer 1

The correct option is B $50$

$wt.ofHCl=\frac{Normality\times Eq.wt\times volume}{1000}wt.ofHCl=\frac{0.1\times 36.5}{5}wt.ofHCl=0.73gm$

Metal carbonate neutralized completely in $HCl$ means number of moles of metal carbonate is equal to No.of moles of $HCl$.

$\frac{wtofmetalcarbonate}{mol.wtofmetalcarbonate}=\frac{wtofHCl}{mol.wtofHCl}$

$mol.wtofmetalcarbonate=\frac{mol.wtofHCl\times wt.ofmetalcarbonate}{wt.ofHCl}$

$mol.wtofmetalcarbonate=\frac{36.5\times 1}{0.73}mol.wtofmetalcarbonate=50.$